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'[PIC]:16f84 Question'
2002\02\04@210910 by Steve Kasm

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Hello All,



On a 16f84, if I bring Mclr to ground will this reset the chip?

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2002\02\04@212313 by Tony Nixon

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Steve Kasm wrote:
>
> Hello All,
>
> On a 16f84, if I bring Mclr to ground will this reset the chip?
>


Yes, but make sure MCLR is tied to VCC via a resistor (10K or so) to
avoid a short.


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2002\02\04@212722 by Andrew Warren

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Steve Kasm <.....PICLISTKILLspamspam.....mitvma.mit.edu> wrote:

> On a 16f84, if I bring Mclr to ground will this reset the chip?

   Yes.

   -Andy

=== Andrew Warren -- EraseMEaiwspam_OUTspamTakeThisOuTcypress.com
=== Principal Design Engineer
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=== Opinions expressed above do not
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=== Cypress Semiconductor Corporation

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2002\02\04@212735 by Jon Baker

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> On a 16f84, if I bring Mclr to ground will this reset the chip?

Yes. Holding it a ground will hold it in a reset state, when it returns to a
logic 1 state the chip will start to execute instructions again.

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2002\02\04@213140 by Jinx

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> On a 16f84, if I bring Mclr to ground will this reset the chip?

Yes, but you risk damage to the chip if you ground it directly

{Quote hidden}

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2002\02\04@215410 by Steve Kasm

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Is there a way to do this using another logic chip to control the reset
or a transistor? I need to control this reset from another signal on my
circuit. I posted a message about this to the [EE] board didn't think it
was appropriate for this board. What I have from this point which is an
inappropriate signal to work with is a rectified dc signal that as logic
high is 41v dc and logic low is 14v dc. Somehow I need that 14v dc to
reset the processor? I need my program to reset at this condition and
I'm out of I/O pins on the processor to deal with this.

Regards,
Steve

{Original Message removed}

2002\02\05@012745 by uter van ooijen & floortje hanneman

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> On a 16f84, if I bring Mclr to ground will this reset the chip?

Yes, that's why it is called /MCLR (/ == active low).

Is this a trick question?

Wouter van Ooijen

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2002\02\05@012757 by uter van ooijen & floortje hanneman

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> Yes, but you risk damage to the chip if you ground it directly

Source for that? I know you can damage it by inrush current (when tied to a
charged cap and no Vcc), but by tying it low???

Wouter

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2002\02\05@033223 by Jinx

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> > Yes, but you risk damage to the chip if you ground it directly
>
> Source for that? I know you can damage it by inrush current (when
> tied to a charged cap and no Vcc), but by tying it low???

I think you may have misunderstood "ground it directly". Not tied
to ground (although why would you want to have the PIC in
permanent reset, but maybe it's my turn to misunderstand), but
as you say, to reduce instantaneous currents. My sources for
this are the F84 manual (yes, I'm back to trusting manuals when
it suits me !!!) and work done, and advice given by, Dan Michaels

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2002\02\05@034728 by Trevor Page

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As for damaging the MCLR input by grounding it directly, I quote from the
datasheet: "Voltage spikes below VSS at the MCLR pin, inducing currents
greater than 80 mA, may cause latch-up. Thus, a series resistor of 50-100ohm
should be used when applying a "low" level to the MCLR pin rather than
pulling this pin directly to VSS".

Okay, in practice you wouldn't want to tie MCLR directly to ground anyway. I
guess they mean if you're pulling it low using a TTL signal, or with a
bipolar transistor / FET. I suppose this advice would apply to any input pin
in general (especially those which can be configured as outputs, of course).


I guess that if you're driving the pin with a TTL signal and/or you've got
some inductance in the circuit (PCB track etc) then MCLR would quite likely
go below Vss. But I wonder if this is really necessary if you're, say,
pulling it low with a simple FET and the track from the FET source to the
PIC's Vss pin is very short?

Having just brushed up on my knowledge of the MCLR input, I'm now worried
that I should have implemented an RC reset in my last PIC -based product
rather than connecting it straight to Vdd: there is a lot of capacitance on
the 5V rail in that circuit. Am I likely to run into trouble in practice?
Comments?

Never stop learning...

Trev


> {Original Message removed}

2002\02\05@042130 by Jinx

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> Having just brushed up on my knowledge of the MCLR input,
> I'm now worried that I should have implemented an RC reset
> in my last PIC -based product rather than connecting it straight
> to Vdd: there is a lot of capacitance on the 5V rail in that circuit.
> Am I likely to run into trouble in practice? Comments?

If you think that there will be voltage stored in the MCLR cap
after Vcc has gone from the PIC then use a signal diode in
parallel with the R (a to MCLR, k to Vcc) to discharge it. Causing
MCLR to rise to Vcc after the Vcc pin itself is no bad thing, but
you could leave out the C and use the R as a pull-up. A link (0R)
is better as an R can allow Vcc glitches and noise through and
possibly cause the PIC to reset

> Never stop learning...

Never stop aksing

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2002\02\05@043921 by Peter Onion

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On 05-Feb-02 Trevor Page wrote:
> As for damaging the MCLR input by grounding it directly, I quote from the
> datasheet: "Voltage spikes below VSS at the MCLR pin, inducing currents
> greater than 80 mA, may cause latch-up. Thus, a series resistor of 50-100ohm
> should be used when applying a "low" level to the MCLR pin rather than
> pulling this pin directly to VSS".

I think the important thing here is to look at the figures.....

I can't see now a simple switch to ground and a 10K pull up to Vcc and a small
capacitor are going to create voltage spikes below ground that can pump
80mA of current into the MCLR pin.  Even with a several inches of wire to the
switch I doubt there would be enough inductance about to cause problems.

Is the MCLR input any different in this respect to other pins ?

Peter

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2002\02\05@055155 by Kevin Olalde

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Does any of this interfere with ISCP?  I have no basis for this question, just
that programming this part has given me troubles.  I don't know if they're due
to me always trying ISCP or the programmer.

I recently bought a ICD so I haven't tried any '628s lately.

Thanks,
Kevin

Jinx wrote:
{Quote hidden}

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2002\02\05@142805 by uter van ooijen & floortje hanneman

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> > > Yes, but you risk damage to the chip if you ground it directly

> I think you may have misunderstood "ground it directly". Not tied
> to ground (although why would you want to have the PIC in
> permanent reset, but maybe it's my turn to misunderstand),

OK, I still can't read 'ground it directly' different from 'tied to ground'
but that might be my problem. Tying MCLR either to ground or Vcc is OK, but
to a current source independent from Vcc (such as a cap) is definitely bad.

And why would anyone tie /MCLR to ground directly? Well, I do that with a
reset switch... When I want a really form reset (of both the PIC and maybe a
HD44780 which does not have a reset input) I tie Vcc to Gnd! With an 7805
that is not a big problem and it cures even a brownout-problem.

Wouter van Ooijen

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2002\02\05@142812 by uter van ooijen & floortje hanneman

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> Having just brushed up on my knowledge of the MCLR input, I'm now worried
> that I should have implemented an RC reset in my last PIC -based product
> rather than connecting it straight to Vdd: there is a lot of capacitance
on
> the 5V rail in that circuit. Am I likely to run into trouble in practice?

The trouble is a current source *independent* from Vcc, that would try to
'power' the PIC via a path that is not up to such a current. So tying it
directly to Vcc or Gnd is no problem, but a cap *is* a problem, so in that
case use a resistor between the cap and the MCLR.

Wouter van Ooijen

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2002\02\05@142815 by uter van ooijen & floortje hanneman

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> I can't see now a simple switch to ground and a 10K pull up to Vcc and a
small
> capacitor are going to create voltage spikes below ground that can pump
> 80mA of current into the MCLR pin.  Even with a several inches of wire to
the
> switch I doubt there would be enough inductance about to cause problems.

The cap can!

> Is the MCLR input any different in this respect to other pins ?

Yes, it does not have the normal protection diode to Vcc (because it must
accept the ~14 V Vpp). I guess RA4 is similar.

Wouter van Ooijen

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2002\02\05@142819 by uter van ooijen & floortje hanneman

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> Does any of this interfere with ISCP?  I have no basis for this question,
just
> that programming this part has given me troubles.  I don't know if they're
due
> to me always trying ISCP or the programmer.

sources of ICSP trouble:
- load on /MCLR (esp. caps), so the rise from vcc to vpp is too slow
- load on any of the involved pins (esp. caps)
- not tying LVP enable low

When you want to use an RC-reset with ICSP split the R: Vcc - 10k - C -
22k - /MCLR with a diode over the 10k.

Wouter van Ooijen

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2002\02\05@143630 by jim

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Wouldn't a cap be a voltage source, not current?

                                               Jim.
{Original Message removed}

2002\02\05@144250 by Roman Black

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Steve Kasm wrote:
>
> Is there a way to do this using another logic chip to control the reset
> or a transistor? I need to control this reset from another signal on my
> circuit. I posted a message about this to the [EE] board didn't think it
> was appropriate for this board. What I have from this point which is an
> inappropriate signal to work with is a rectified dc signal that as logic
> high is 41v dc and logic low is 14v dc. Somehow I need that 14v dc to
> reset the processor? I need my program to reset at this condition and
> I'm out of I/O pins on the processor to deal with this.


Sure, use two voltage dividers (4 resistors total)
and a logic chip, quad nand gate etc. Each voltage
divider is different, and each goes to a input pin.
At 30v+ dc both inputs are high,
at 14v dc one one logic gate is high.
Configure the gates with whatever chip you prefer
so that it generates a low output when the input
is 14v. :o)
-Roman

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2002\02\05@160708 by Jinx

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> OK, I still can't read 'ground it directly' different from 'tied to
ground'

Think we're talking about two completely different things. I interpret
"ground it directly" as using a low-ohms device (switch) to connect
MCLR to ground so as to reset the PIC. After the switch is released,
MCLR goes back to being held high via a pull-up. I interpret "tied to
ground" as a permanent ohmic connection, eg a wire link or PCB
trace, that holds MCLR at 0V forever and ever amen (the opposite
of the more usual "tying" MCLR to Vcc via a trace when no reset
facility other than power-down-power-up is required). Permanently
holding the PIC in reset by "tying" MCLR down didn't make sense,
but it seems we were both victims of semantics

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2002\02\06@052802 by Peter Onion

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On 05-Feb-02 wouter van ooijen & floortje hanneman wrote:
>> I can't see now a simple switch to ground and a 10K pull up to Vcc and a
> small
>> capacitor are going to create voltage spikes below ground that can pump
>> 80mA of current into the MCLR pin.  Even with a several inches of wire to
> the
>> switch I doubt there would be enough inductance about to cause problems.
>
> The cap can!

Wouter,

Please explain in more detail how you think this can happen.

Do you mean that the capacitor (acting as a capacitor) can create the negative
voltage, OR that the capacitor (acting as an inductor) can create the negative
voltage.
>
>> Is the MCLR input any different in this respect to other pins ?
>
> Yes, it does not have the normal protection diode to Vcc (because it must
> accept the ~14 V Vpp). I guess RA4 is similar.

OK, but we are talking about problems caused by negative spikes going below Vss
not posititive spikes going above Vcc.

Peter

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2002\02\06@053856 by Peter Onion

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On 25-Jun-07 jim wrote:
> Wouldn't a cap be a voltage source, not current?

The charge stored within a capacitor gives rise to the voltage across it.
If you put an external load on a charged capacitor the voltage will cause a
current to flow in the load.  The loss of stored charge causes the voltage to
drop.

Maybe you are thinking of constant current source rather than constant voltage
source ?   The capacitor is neiter of these of course !

Peter

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2002\02\06@100033 by James Paul

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Peter,

I am just saying it would supply a voltage, not a current as capacitors
are voltage devices.   And as an FYI, I didn't say it would be a constant
anything source.  Just that it is more a voltage source than a current
source.   I guess it really doesn't matter.  It was just an observation.

                                                    Thanks and Regards,

                                                           Jim



{Quote hidden}

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2002\02\06@102045 by Peter Onion

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On 06-Feb-02 James Paul wrote:
>
>  Peter,
>
>  I am just saying it would supply a voltage, not a current as capacitors
>  are voltage devices.

I don't think you can classify capacitors as either a voltage OR a current
device, atleast not in the same way you might call a bipolar transisor a current
device but a FET a voltage device (actually its a transconductance device if I
remember correctly).

>  And as an FYI, I didn't say it would be a constant anything source.

I know you didn't, and no where did I say you did.  Mind you, rereading my post
I didn't make my point very clearly.

>  Just that it is more a voltage source than a current source.
> I guess it really doesn't matter.  It was just an observation.

Whether you consider it to be acting as a current or voltage source depends
on how you chose to model it and the circuit it is connected to.

Look up Norton & Thevenin equivalent circuits and you will see that voltage and
current sources are interchangeable.

Peter.

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2002\02\06@151212 by Roman Black

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James Paul wrote:
>
>  Peter,
>
>  I am just saying it would supply a voltage, not a current as capacitors
>  are voltage devices.   And as an FYI, I didn't say it would be a constant
>  anything source.  Just that it is more a voltage source than a current
>  source.   I guess it really doesn't matter.  It was just an observation.
>


Having stuck screwdrivers across the terminals
of thousands of capacitors, they sure seem like
a current source to me! ;o)
-Roman

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